Physics 2101 Section 3 March 12 : Ch. 10 - LSUjzhang/Lecture19_2101.pdf · Physics 2101 Section 3...
Transcript of Physics 2101 Section 3 March 12 : Ch. 10 - LSUjzhang/Lecture19_2101.pdf · Physics 2101 Section 3...
Physics 2101 Physics 2101 Section 3Section 3Section 3Section 3
March 12March 12rdrd : Ch. 10: Ch. 10
Announcements:Announcements:•• MidMid--grades posted in PAWgrades posted in PAW•• Quiz todayQuiz today•• I will be at the March APS I will be at the March APS meeting the week of 15meeting the week of 15--1919thth. Prof. Rich Kurtz will . Prof. Rich Kurtz will help mehelp mehelp me.help me.
Class Website: http://www.phys.lsu.edu/classes/spring2010/phys2101http://www.phys.lsu.edu/classes/spring2010/phys2101--3/3/http://www.phys.lsu.edu/~jzhang/teaching.htmlhttp://www.phys.lsu.edu/~jzhang/teaching.html
Forces of Rolling
1) If object is rolling with a =0 (i.e. no net forces), then v =ωR = constant (smooth roll)1) If object is rolling with acom 0 (i.e. no net forces), then vcom ωR constant (smooth roll)
…if constant speed, it has no tendency to slide at point of contact ‐ no frictional forces
a = 0 α = 0 τ = 0 f = 0acom 0 α 0 τ 0 f 0
acom = αR Icomα = τnet τnet = |R||f|sinθ
f d d l !
2) If object is rolling with acom≠ 0 (i.e. there are net forces) and no slipping occurs,then α ≠ 0 ⇒ τ ≠ 0
… static friction needed to supply torque !
kinetic friction - during sliding (acom ≠ αR)
t ti f i ti th lli ( R)static friction - smooth rolling (acom = αR)
Rolling down a ramp : Energy considerations
An object at rest with massm and radius r
h
An object at rest, with mass m and radius r, roles from top of incline plane to bottom. What is v at bottom and a throughout.
ΔE 0θ
L sinθ = h
KE( ) 0( ) = 0( ) mgh( )[ ]
ΔEmech = 0ΔKEtot = −ΔU
KErot + trans,COM( )final− 0( )init
= − 0( ) final− mgh( )init[ ]
12 mvCOM
2 + 12 ICOM ω 2 = mgL sinθ
Using v2 = v2 + 2aLUsing vcom = ωr
v =2gL sinθ Using 1‐D
Using v = v0 + 2aL
a =v2
=gsinθ
⎛ ⎞vCOM =1+
Icom
mR2
⎛ ⎝ ⎜
⎞ ⎠ ⎟ kinematics
(const accel)
acom =2L
=1+
Icom
mR2
⎛ ⎝ ⎜
⎞ ⎠ ⎟
Speedat Bottom
ConstantAcceleration
Rolling down a ramp: acceleration‐ Acceleration downwards
‐ Friction provides torque
‐ Static friction points upwardsYo‐Yo
ˆ x : fs − Fg sinθ = −macom
ˆ θ
Newton’s 2nd LawLinear version
ˆ y : N − Fg cosθ = 0
ˆ z : Icomα = τ = Rfs
acom =R2 mgsinθ − macom( )
I
Angular version‐ Tension provides torque‐ Here θ = 90°A l R R
→ acom = αR
α =acom
R=
Rfs
Icom
com Icom ‐ Axle: R ⇒ R0
acom =g
1 I R2com
acom =gsinθ
1+ Icom mR2
com 1+ Icom mR02
Consider a round uniform body of mass and radius M RRolling Down a Ramp
acomcom
rolling down an inclined plane of angle . We willcalculate the acceleration of the center of mass along the a is sing
aθ
Ne ton's second la for thealong the -axis using x Newton's second law for the translational and rotational motion.
com
com
Newton's second law for motion along the -axis: sin (eq. 1)Newton's second law for rotation about the center of mass:
s
s
x f Mg MaRf I
θτ α
− == = com
com We substitute in the second equation and g
sfaR
α α= − comcomet s
aRf IR
a
= − →
comcom 2
com
(eq. 2). We substitute from equation 2 into equation 1
sin
s saf I fR
aI Mg Maθ
= − →
− − = comsinga I
θ= −com com2 sinI Mg Ma
Rθ com
com21 I
MR+
comsin| | ga I
θ=
acom
comcom
21 IMR
+
Cylinder Hoop2
21 2
2sin sin
MRI I MR
g gθ θ
= =
1 22 21 2
1
sin sin 1 / 1 /
sin
g ga aI MR I MR
ga
θ θ
θ
= =+ +
= 2singa θ
=1 21 / 2a
MR+ 22 2 2
1 2
1 /
sin sin 1 1/ 2 1 1
aMR MR MR
g ga aθ θ+
= =+ +
1 22 sin sin(0.67) sin (0.5) sin
3 2g ga g a gθ θθ θ= = = =
Prob. 11‐64 NON‐smooth rolling motion
A bowler throws a bowling ball of radius R along a lane. The ball slides on the lane, with initial speed vcom,0 and initial angular speed ω0 = 0. The coefficient of kinetic friction between the ball and the lane is μk. The kinetic frictional force fk acting on the ball while producing a torque that causes an angular acceleration of the ball. When the speed vcom has decreased enough and the angular angular acceleration of the ball. When the speed vcom has decreased enough and the angular speed ω has increased enough, the ball stops sliding and then rolls smoothly.
a) [After it stops sliding] What is the vcom in terms ofω ?
b) During the sliding, what is the ball’s linear acceleration?Smooth rolling means vcom =Rω
) g g,
c) During the sliding, what is the ball’s angular acceleration?
ˆ x : − fk = macom
ˆ y : N − mg = 0From 2nd law:(linear)
But fk = μ k N= μ k mg
acom = − fk m= −μ k g
So
) g g, g
d) What is the speed of the ball when smooth rolling begins?
τ = Rfk (− ˆ z )Iα(− ˆ z ) =
τ = Rfk (− ˆ z )
From 2nd law:(angular)
But fk = μ k N= μ k mg
So Iα = Rfk = R μ k mg( )
α = Rμ k mgI
⎛ ⎞ ⎛ ⎞ ) p g g
When does vcom =Rω ? vcom = v0 + acom tω = ω 0 + αtFrom kinematics:
vcom = v0 − μ k gt
t = ωα = Iω
Rμ k mgvcom = v0 − μ k g
I vcomR
⎛ ⎝ ⎜ ⎞
⎠ ⎟
Rμ k mg
⎛
⎝
⎜ ⎜ ⎜
⎞
⎠
⎟ ⎟ ⎟
e) How long does the ball slide? vcom =v0
1+ I mR2( )t =
v0 − vcom
μ k g
Torque & Angular Momentumω
Conservation of:( )• Linear momentum (closed, isolated system)
• KE (elastic collisions)• Mechanical energy (only conservative forces)• Total energy
• Angular momentum (closed, isolated system)
Angular Momentum(single particle)
Linear momentumwhen particle passes through point A( g p ) Ait has linear momentum:
p = mv
with components in x‐y planeAngular momentumwith respect to point O the angular momentumis defined as:
( ) r where is the position vector of the particle
with respect to O.
= r × p = m r × v ( )
p
•Units kg⋅m2/s•components in z‐direction only !
(⊥ to x‐y plane) •just as in has meaning only with respect to a specified point (axis)
τ ,
Newton’s 2nd Law in Angular Form(single particle)
Relationship between force and linear momentum d
F net = F i∑ = m a = dp
dtNewton’s 2nd Law in linear form
Relationship between torque and angular momentum
∑ d The vector sum of all the torques acting on a
τ net = τ i∑ =
ddt
q gparticle is equal to the time rate change of the angular momentum of that particle.
has no meaning unless the net torque , and the rotational momentum , are defined with respect to the same origin
τ net = d dt
τ net
, p g
Example #1A cannon ball, with mass m, is shot directly upward with velocity v0. What is the angular momentum and torque of the ball about a point a distance x0 from the cannon?
The only force on the in‐flight cannon ball is gravity:
Th t iti d l it f th i fli ht b ll i F g = −mgˆ j
The vector position and velocity of the in‐flight ball is: ˆ x : x(t) = x0 + v0,x t = x0
ˆ y : y(t) = y0 + v0,y t + 12 at 2 = v0t − 1
2 gt 2
r = x0( )ˆ i + v0t − 12 gt 2( )ˆ j
v = v0 − gt( )ˆ j
The vector angular momentum about the man of the in‐flight ball is:
l = mr ×v = m
ˆ i ˆ j ˆ k x0 v0t − 1
2 gt 2 0 = mx0 v0 − gt( )ˆ k 0 v0 − gt 0
The vector torque about the man of the in‐flight ball is: ˆ i ˆ j ˆ k
Check
τ = d
l dt
=ddt
mx0 v0 − gt( ) ˆ k [ ]k
τ = r × F =i j k x0 v0t − 1
2 gt 2 00 −mg 0
= −x0mg ˆ k = −mx0g k
Newton’s 2nd Law in Angular FormSystem of particles
Total angular momentum n
Relationship between torque and angular momentum for system of particles L = l 1 + l 2 + l 3 + ....+ l n = l i
i=1
n
∑
τ net =dL dt τ net represents the net external torque.
Example #2(Problem 11‐27)
#1( )
Two particles are moving as shown. a) What is their total moment of inertia about point O?b) What is their total angular momentum about point O?
#2b) What is their total angular momentum about point O?c) What is their net torque about point O?
a) The total moment of inertia about point O is found from: I I∑ m r 2∑ (6 5 kg)(1 5 m)2 + (3 1 kg)(2 8 m)2 38 9 kg m 2
b) The total angular momentum about point 0 is:
Itot = Ii∑ = miri∑ = (6.5 ⋅ kg)(1.5 ⋅ m) + (3.1⋅ kg)(2.8 ⋅ m) = 38.9 ⋅ kg ⋅ m
L =rl i∑ = mi
rri ×rvi( )∑ = m1 r1v1 sinθ1( )(− z) + m2 r2v2 sinθ2( )(+ z) Ltot l i∑ mi ri × vi( )∑ m1 r1v1 sinθ1( )( z) + m2 r2v2 sinθ2( )(+z)
L tot = −(6.5 ⋅ kg)(1.5 ⋅ m)(2.2 ⋅ m /s) + (3.1⋅ kg)(2.8 ⋅ m)(3.6 ⋅ m /s)[ ]ˆ z
= 9.8 ⋅ kg ⋅ m 2 /s ⋅ ˆ z
c) The net torque about point 0 is:
dL tot closed isolated system0
τ net = tot
dtclosed, isolated system= 0
Angular Momentum of a Rigid BodyRotating about a Fixed axis
τ Iα dL I dω τ net = Iα
L I
dt= = I
dt
E l #3 (P bl 11 39)
L = Iω
Example #3 (Problem 11‐39) Three particles are fastened to each other with massless string and rotate around point O.What is the total angular momentum about point O?
Itot = m d( )2 + m 2d( )2 + m 3d( )2
14 d 2= 14md 2
L = Iω = 14mωd 2 ⋅ z L tot = Iω = 14mωd z [kg⋅m2/s]